A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 35.0 m above sea level, directed at an angle theta = 45.9° above the horizontal, and with a speed v = 32.3 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.

Vo = 32.3m/ @ 45.9o.

Xo = 32.3*cos45.9 = 22.5 m/s.
Yo = 32.3*sin45.9 = 23.20 m/s.

Y = Yo + g*t.
Tr = (Y-Yo)/g = (0-23.2)/-9.8 = 2.37 s.
= Rise time.

Tf1 = Tr = 2.37 s. = Fall time from hmax
to 35 m above sea level.

h = Yo*t + 0.5g*t^2 = 35 m.
23.2*t + 4.9t^2 = 35
4.9t^2 + 23.2t - 35 = 0.
Use Quadratic Formula and get:
Tf2 = 1.20 s. = Fall time from 35 m above sea level to sea level.

T = Tr + Tfi + Tf2.
T = 2.37 + 2.37 + 1.20 = 5.94 s. = Time
in air.

D = Xo * T = 22.5m/s * 5.94s = 133.7 m.

To find the horizontal distance traveled by the projectile, we can use the equations of projectile motion. The horizontal and vertical motions are independent of each other.

First, let's find the time of flight (T) of the projectile. The time it takes for the projectile to reach the same vertical height on its way up and on its way down is equal to half of the total time of flight. We can find this time by using the equation:

H = (1/2) * g * T^2

where H is the vertical displacement (35.0 m) and g is the acceleration due to gravity (9.8 m/s^2).

Rearranging the equation to solve for T, we have:

T^2 = (2 * H) / g

T^2 = (2 * 35.0) / 9.8

T^2 ≈ 7.14

T ≈ √7.14

T ≈ 2.67 s (rounded to two decimal places)

Now, we can find the horizontal distance traveled by the projectile using the equation:

D = v * T * cos(theta)

where v is the initial velocity (32.3 m/s), T is the time of flight (2.67 s), and theta is the launch angle (45.9°).

Plugging in the values we know, we have:

D = (32.3 m/s) * (2.67 s) * cos(45.9°)

D ≈ 91.30 m (rounded to two decimal places)

Therefore, the horizontal distance traveled by the projectile is approximately 91.30 meters.

To find the horizontal distance traveled by the projectile, you can use the equation:

D = (v^2 * sin(2*theta)) / g

where:
D is the horizontal distance traveled
v is the initial velocity of the projectile
theta is the launch angle
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Given that:
v = 32.3 m/s
theta = 45.9° (convert to radians: theta = 45.9° * π/180°)
g = 9.8 m/s^2

We can plug these values into the equation:

D = (32.3^2 * sin(2*(45.9° * π/180°))) / 9.8

Calculating the value of sin(2*(45.9° * π/180°)) gives approximately 0.874, so we have:

D = (32.3^2 * 0.874) / 9.8

Simplifying further:

D = (1051.29 * 0.874) / 9.8

D = 92.06 m

Therefore, the horizontal distance traveled by the projectile is approximately 92.06 meters.